If the antiderivative of f(x) =sin⁡xsin2⁡x+4cos2⁡x is (1/3)tan−1⁡((1/3)g(x))+C theng(x) is equal to

f(x)=sin⁡xsin2⁡x+4cos2⁡x=tan⁡xsec⁡xtan2⁡x+4=tan⁡xsec⁡xsec2⁡x+3Putting sec⁡x=t,dxsec⁡xtan⁡x=dt so∫f(x)dx=∫dtt2+3=13tan−1⁡sec⁡x3+C